A flatness result of Fiedorwicz for amalgamated free products of monoids in connection with classifying spaces of monoids In Lemma 5.2(a) of Z. Fiedorowicz, Classifying Spaces of Topological Monoids and Categories American Journal of Mathematics Vol. 106, No. 2 (Apr., 1984), pp. 301-350 the author proves the following.

Lemma 5.2(a) Suppose that $\{M_i\}_{i\in I}$ is a collection of monoids with a common submonoid $W$ such that the monoid ring $\mathbb ZM_i$ is flat as a left $\mathbb ZW$-module for each $i\in I$.  Let $M$ be the amalgamated free product (or pushout) of the $M_i$ over $W$.  Then $\mathbb ZM$ is flat as a left $\mathbb ZM_j$-module for each $j\in I$.

His proof is one line, but I cannot understand it.  I would very much love to use this result.  I'm a bit nervous about it because the proof he sketches would seem to work in the category of rings, or more generally $k$-algebras over a commutative ring with unit $k$, but papers I saw of Cohn on amalgamated free products of rings seem to involve much more complicated colimits diagrams than Fiedorwicz is using.  Moreover, Warren Dicks pointed out to me an example of an amalgamated free product of $k$-algebras, with $k$ a field, where the factors are flat over the subring but the amalgamated free product is not flat over the factors.  He derives it from the paper https://arxiv.org/abs/math/0205034.
I would be happy for a proof I can understand in the case of two factors.  I did try to contact the author, but he is retired and didn't respond.  I now reproduce his proof.

Proof. This is immediate because $\mathbb ZM$ is the direct limit of
$$\mathbb ZM_j\otimes_{\mathbb ZW} \mathbb ZM_{i_1}\otimes_{\mathbb ZW} \mathbb ZM_{i_2}\otimes_{\mathbb ZW}\cdots \otimes_{\mathbb ZW}\mathbb ZM_{i_k}$$ $i_1,\ldots, i_k\in I$ as a left $\mathbb ZM_j$-module and hence is flat over $\mathbb ZM_j$.

What I don't understand is exactly what is the directed (or filtered) diagram that this direct limit is indexed over (i.e., what are the maps). Each of the terms is flat, so if $\mathbb ZM$ is such a direct limit, then I am fine with the proof.  There is a map from each of these iterated tensors to $\mathbb ZM$ induced by multiplication (and the ``inclusions'' into the pushout) but I don't see how to organize this into a direct limit (=filtered colimit).
 A: This appears to be false to me (unless maybe there are commutativity hypotheses), unless I've made a mistake below. (This is based on an example that I saw Andrew Ranicki give, a number of years ago, about non-exactness of Cohn localization.)
Consider the diagram of monoids
$$
\Bbb Z \leftarrow \Bbb N \rightarrow \Bbb N \ast \Bbb N
$$
where the right-hand arrow is the inclusion of the first factor into the free product.  On taking monoid algebras, this is a diagram of rings
$$
\Bbb Z[x^{\pm 1}] \leftarrow \Bbb Z[x] \rightarrow \Bbb Z\langle x,y\rangle.
$$
The left factor is clearly flat over $\Bbb Z[x]$ and the right factor is free on the basis $$\{1\} \cup\{y\cdot w \mid w \text{ is a monomial in }x,y\}.$$
Thus both factors are flat left modules. The monoid algebra on the pushout $M$ is the ring $\Bbb Z\langle x^{\pm 1}, y\rangle$ by a universal property argument.
However, consider the exact sequence of left $\Bbb Z\langle x,y\rangle$-modules
$$
0 \to \Bbb Z\langle x,y\rangle \oplus \Bbb Z\langle x,y\rangle \xrightarrow{[x,y]} \Bbb Z\langle x,y\rangle \to \Bbb Z \to 0,
$$
where the first map sends $(a,b)$ to $ax+by$ and the second sends $x$ and $y$ to zero. Tensoring on the left with $\Bbb Z\langle x^{\pm 1}, y\rangle$ gives the sequence
$$
0 \to \Bbb Z\langle x^{\pm 1},y\rangle \oplus \Bbb Z\langle x^{\pm 1},y\rangle \xrightarrow{[x,y]} \Bbb Z\langle x^{\pm 1}, y\rangle \to 0 \to 0.
$$
This is not left exact, because the first factor of the direct sum is taken isomorphically to the middle term by invertibility of $x$.
